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7.6 Entropy Change in Irreversible ProcessesPowerPoint Presentation

7.6 Entropy Change in Irreversible Processes

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7.6 Entropy Change in Irreversible Processes.

7.6 Entropy Change in Irreversible Processes

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- It is not possible to calculate the entropy change ΔS = SB - SA for an irreversible process between A and B, by integrating dq / T, the ratio of the heat increment over the temperature, along the actual irreversible path A-B characterizing the process.
- However, since the entropy is a state function, the entropy change ΔS does not depend on the path chosen.The calculation of an irreversible process can be carried out via transferring the process into many reversible ones:
- Three examples will be discussed here: (1) heat exchange between two metal blocks with different temperatures; (2) Water cooling from 90 to a room temperature; (2) A falling object.

Thermodynamics Potential

Chapter 8

- Thermodynamic potentials: Helmholtz function F and the Gibbs function G.
- The enthalpy, Helmholtz function and Gibbs functions are all related to the internal energy and can be derived with a procedure known as Legendre differential transformation.
- The combined first and second laws read
dU = Tds – PdV

where T and S, and -P and V are said to be canonically conjugate pairs.

- By assuming U = U(S,V), one has

- Consider a function Z = Z(x, y), the differential equation is dZ = Xdx + Ydy
where X and x, Y and y are by definition canonically conjugate pairs.

- We wish to replace (x, y) by (X, Y) as independent variables. This can be achieved via transforming the function Z(x,y) into a function M(X,Y).
- Assume M(X,Y) = Z(x,y) – xX – yY
Then dM = dZ – Xdx – xdX –Ydy – ydY

dM = -xdX - ydY

- The change in internal energy is the heat flow in an isochoric reversible process.
- The change in enthalpy H is the heat flow in an isobaric reversible process.
- The change in the Helmholtz function in an isothermal reversible process is the work done on or by the system.
- The decrease in F equals the maximum energy that can be made available for work.

- Based on the second law of thermodynamics
dQ ≤ T∆S with dQ = ∆U + P ∆V

- Combine the above expressions
∆U + P ∆V ≤ T∆S

∆U + P ∆V - T∆S ≤ 0

- Since G = U + PV –TS
(∆G)T,P≤ 0 at constant T and P

or G f ≤ Gi

- Gibbs function decreases in a process until a minimum is reach, i.e. equilibrium point.
- Note that T and P need not to be constant throughout the process, they only need to have the same initial and final values.